1. Ta có $0\le\sin^2x;\cos^2x\le 1$, suy ra:
$15\cos^2x\le 1993\cos^2x$
$1993\sin^{1992}x\le1993\sin^2x$
$\Rightarrow 15\cos^2x+1993\sin^{1992}x\le1993(\sin^2x+\cos^2x)=1993$
Dấu bằng xảy ra khi $\cos x=0 \Leftrightarrow x=\dfrac{\pi}{2}+k\pi,k\in\mathbb{Z}$